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Cylindrical tank

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Cylindrical tank

by rahulvsd » Mon Mar 05, 2012 7:35 am
A cylindrical tank with radius 3 meters is filled with a solution. The volume at t minutes is given by V = (-t^2 + 12t + 24)Ï€ m3. At 4 minutes, there are d meters of unused vertical space in the tank. At 6 minutes, there are only 1/3d meters of space remaining. What is the height of the tank?
Choices
A 6*(5/9)

B 6*(8/9)

C 7 *(1/3)

D 7* (5/9)

E 8

[spoiler]Ans: B[/spoiler]
Last edited by rahulvsd on Mon Mar 05, 2012 6:26 pm, edited 1 time in total.
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by Neo Anderson » Mon Mar 05, 2012 10:42 am
dear friend, check the question you have typed, seems to be a typo, cause assuming -t2 in
V = (-t2 + 12t + 24)Ï€ m3
is -(t)^2 and also
1/3d
is d/3, i get the height as 62/9 m which is not the option. here goes the solution:-

Volume of a cylinder is pie r^2 h = 9 x pie x h

from V@4 = (-(4)^2 + 12x4 +24)pie = 56 pie or 9 x pie x (56/9) => h - 56/9 = d --- (1)
from V@6 = (-(6)^2 + 12x6 +24)pie = 60 pie or 9 x pie x (60/9) => h - 60/9 = d/3 --- (2)

from (1) and (2) you get d = 2/3 and hence h = 62/9

pls do check and confirm
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by rahulvsd » Mon Mar 05, 2012 6:26 pm
Sorry typo error. Please read that as t square.
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